Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const pair<int, int> DD[] = {{1, 0},  {1, 1},   {0, 1},  {-1, 1},
                             {-1, 0}, {-1, -1}, {0, -1}, {1, -1}};
template <class T>
T smin(T &a, T b) {
  if (a > b) a = b;
  return a;
}
template <class T>
T smax(T &a, T b) {
  if (a < b) a = b;
  return a;
}
int main(void) {
  srand(time(0));
  cout << fixed << setprecision(7);
  cerr << fixed << setprecision(7);
  static int v[1 << 10][1 << 10];
  static int s[1 << 10][1 << 10];
  vector<int> row, col;
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= m; ++j) cin >> v[i][j];
  int ok = 0;
  if (n > m) {
    swap(n, m);
    ok = 1;
    for (int i = 1; i <= n; ++i)
      for (int j = 1; j <= m; ++j) s[i][j] = v[j][i];
  } else {
    for (int i = 1; i <= n; ++i)
      for (int j = 1; j <= m; ++j) s[i][j] = v[i][j];
  }
  for (int i = 1; i <= n; ++i) {
    int mn = 1e9;
    for (int j = 1; j <= m; ++j) smin(mn, s[i][j]);
    for (int k = 0; k < mn; ++k) row.push_back(i);
    for (int j = 1; j <= m; ++j) s[i][j] -= mn;
  }
  for (int j = 1; j <= m; ++j) {
    int mn = 1e9;
    for (int i = 1; i <= n; ++i) smin(mn, s[i][j]);
    for (int k = 0; k < mn; ++k) col.push_back(j);
    for (int i = 1; i <= n; ++i) s[i][j] -= mn;
  }
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= m; ++j)
      if (s[i][j] != 0) return puts("-1") * 0;
  cout << (row.size() + col.size()) << '\n';
  for (auto it : row) cout << (!ok ? "row " : "col ") << it << '\n';
  for (auto it : col) cout << (!ok ? "col " : "row ") << it << '\n';
  cerr << "Time elapsed :" << clock() * 1000.0 / CLOCKS_PER_SEC << " ms"
       << '\n';
  return 0;
}